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3 years ago

Simplify (3x2y)(5x3y−1)

Mathematics

1 answer:

ludmilkaskok [199]3 years ago

3 0

Answer:

6xy(15xy−1)

Step-by-step explanation:

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PLEASE HELO ASAP WILL GIVE BRAINLEAST TO RIGHT ANSWER

marishachu [46]

Answer:

1/4

Step-by-step explanation:

Favorable Outcomes/ All Possible Outcomes= 2/8

2/8= 1/4

The answer is 1/4.

5 0

4 years ago

Which of these situations is represented by 24 divided by 6

sveticcg [70]

We need to see the answer choices.

3 0

3 years ago

Susan collected 89 stickers for 28 weeks. How many stickers did she collect in total round off your answer to the nearest hundre

Temka [501]

Answer:

2500 ( nearest hundred)

Step-by-step explanation:

Given that :

Number of stickers collected per week = 89

Number of weeks for which stickers were being collected = 28

To obtain the total number of stickers collected, we take the product of the number of weeks for which the stickers were beug vcilled and the number of stickers collected per week

The number of stickers collected :

89 * 28 = 2492

2492 = 2500 ( nearest hundred)

7 0

3 years ago

Melody wants to buy a present for her friend for her birthday. She wants to purchase a CD that costs $18. She thought she had $2

miv72 [106K]

Answer: 20%

Step-by-step explanation: $20-$16=$4. 4/20=1/5 or 20%.

6 0

3 years ago

Read 2 more answers

The weights of broilers (commercially raised chickens) are approximately normally distributed with mean 1387 grams and standard

yuradex [85]

Answer:

Probability that a randomly selected broiler weighs more than 1454 g is 0.3372 or 34% (approx.)

Step-by-step explanation:

Given:

Weights of Broilers are normally distributed.

Mean = 1387 g

Standard Deviation = 161 g

To find: Probability that a randomly selected broiler weighs more than 1454 g.

we have ,

$Mean,\,\mu=1387$

$Standard\,deviation,\,\sigma=161$

X = 1454

We use z-score to find this probability.

we know that

$z=\frac{X-\mu}{\sigma}$

$z=\frac{1454-1387}{161}=0.416=0.42$

P( z = 0.42 ) = 0.6628 (from z-score table)

Thus, P( X ≥ 1454 ) = P( z ≥ 0.42 ) = 1 - 0.6628 = 0.3372

Therefore, Probability that a randomly selected broiler weighs more than 1454 g is 0.3372 or 34% (approx.)

8 0

3 years ago